%I #22 Aug 07 2015 03:48:59
%S 1,1,1,1,4,2,1,13,18,6,1,40,116,96,24,1,121,660,1020,600,120,1,364,
%T 3542,9120,9480,4320,720,1,1093,18438,74466,121800,94920,35280,5040,1,
%U 3280,94376,576576,1394064,1653120,1028160,322560,40320,1,9841,478440,4319160
%N Triangle read by rows, k!*2^k*S_2(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.
%C The Stirling-Frobenius subset numbers are defined in A225468 (see also the Sage program).
%H Vincenzo Librandi, <a href="/A225476/b225476.txt">Rows n = 0..50, flattened</a>
%H Peter Luschny, <a href="http://www.luschny.de/math/euler/GeneralizedEulerianPolynomials.html">Generalized Eulerian polynomials.</a>
%H Peter Luschny, <a href="http://www.luschny.de/math/euler/StirlingFrobeniusNumbers.html">The Stirling-Frobenius numbers.</a>
%H Shi-Mei Ma, Toufik Mansour, Matthias Schork, <a href="http://arxiv.org/abs/1308.0169">Normal ordering problem and the extensions of the Stirling grammar</a>, Russian Journal of Mathematical Physics, 2014, 21(2), arXiv 1308.0169 p. 12.
%F T(n, k) = sum_{j=0..n} A_2(n, j)*binomial(j, n-k), where A_2(n, j) are the generalized Eulerian numbers of order m=2.
%F For a recurrence see the Maple program.
%e [n\k][0, 1, 2, 3, 4, 5 ]
%e [0] 1,
%e [1] 1, 1,
%e [2] 1, 4, 2,
%e [3] 1, 13, 18, 6,
%e [4] 1, 40, 116, 96, 24,
%e [5] 1, 121, 660, 1020, 600, 120.
%p SF_SSO := proc(n, k, m) option remember;
%p if n = 0 and k = 0 then return(1) fi;
%p if k > n or k < 0 then return(0) fi;
%p k*SF_SSO(n-1, k-1, m) + (m*(k+1)-1)*SF_SSO(n-1, k, m) end:
%p seq(print(seq(SF_SSO(n, k, 2), k=0..n)), n = 0..5);
%t EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n - k) + m - 1)*EulerianNumber[n - 1, k - 1, m] + (m*k + 1)*EulerianNumber[n - 1, k, m]]); SFSSO[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n - k], {j, 0, n}]/m^k; Table[ SFSSO[n, k, 2], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, May 29 2013, translated from Sage *)
%o (Sage)
%o @CachedFunction
%o def EulerianNumber(n, k, m) :
%o if n == 0: return 1 if k == 0 else 0
%o return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m)+(m*k+1)*EulerianNumber(n-1, k, m)
%o def SF_SSO(n, k, m):
%o return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))/m^k
%o for n in (0..6): [SF_SSO(n, k, 2) for k in (0..n)]
%Y T(n, 0) ~ A000012; T(n, 1) ~ A003462; T(n, 2) ~ A007798.
%Y T(n, n) ~ A000142; T(n, n-1) ~ A001563.
%Y Alternating row sum ~ A000364 (Euler secant numbers).
%Y Cf. A225468, A131689 (m=1).
%K nonn,tabl
%O 0,5
%A _Peter Luschny_, May 19 2013